3. The attempt at a solution
The only information given in the problem is the 2 angles of refraction and of course n of air is 1, I do not understand how to get the index of refraction for both wavelengths with just the angles given and nothing more. They do not state the wavelengths of the rays. I assume it is solvable because it is in the text but if someone could maybe lead me in any direction I would appreciate it.
Thanks

It's not asking for their wavelengths, you just have to assume the wavelengths are different so that the refractive index can be different.
It's simply a geometry question, find the angle from the normal for each of the exit rays and then just use Snell's law to calculate 'n' for each of them.

It's not asking for their wavelengths, you just have to assume the wavelengths are different so that the refractive index can be different.
It's simply a geometry question, find the angle from the normal for each of the exit rays and then just use Snell's law to calculate 'n' for each of them.

Thats what I was thinking so you will have na*sin(theta of a) equals to both nb*sin(0b) and nc*sin(0c) (b and c the 2 refractions) so if u set the 2 equal to each other:
nb*sin(20.5) = nc*sin(12)
dont you need more info to get the exact values instead of just a ratio?
thanks for the quick response

And the angle from normal inside the prism is just the angle from normal to the horizontal dashed line outside ( and is equal to the prism wedge angle)
na sin(x) = sin (12 + x)
nb sin(x) = sin(20.5 + x)
It should be possible to rearrange this to get exact values of na and nb without knowingx